☰ Contents · Algebra

Solving a quadratic inequality by the graph

Lessons 8 · 1 lessons · Sh.A. Alimov, A.R. Xalmukhamedov, M.A. Mirzaakhmedov. Algebra: textbook for Grade 9 of general secondary schools. Revised 4th edition. Tashkent: “O‘qituvchi”, 2019
8

Solving a quadratic inequality with the graph of a quadratic function

Textbook: pp. 28–31
GoalSolve a quadratic inequality using a sketch of the graph of the quadratic function; distinguish the cases D > 0, D = 0, D < 0.
New words
sketch of the graph · grafikning eskizipoint of tangency · urinish nuqtasistrict and non-strict inequality · qat’iy va noqat’iy tengsizlikdirection of the branches · parabola tarmoqlarining yo‘nalishi
Explanation

To solve ax² + bx + c > 0 or < 0 we read the sign of y = ax² + bx + c from its graph. Procedure: 1) the sign of a gives the direction of the branches; 2) find the real roots of ax² + bx + c = 0 or see that there are none; 3) using the points where the parabola crosses (or touches) Ox, draw a sketch; 4) read off the part with the required sign. If D > 0 the parabola crosses Ox at two points; if D = 0 it touches Ox at one point; if D < 0 it does not meet Ox and lies entirely above it (a > 0) or below it (a < 0). For a strict inequality (>, <) the zeros are not solutions, for a non-strict one (≥, ≤) they are.

Worked examples
2x² − 5x − 3 ≤ 0: the branches point up; D = 25 + 24 = 49, x = (5 ± 7) : 4, so x₁ = −1/2, x₂ = 3. The part of the parabola on or below Ox: −1/2 ≤ x ≤ 3.
x² − 6x + 9 = (x − 3)²: the branches point up, D = 0, and the parabola touches Ox at (3, 0). The solution of x² − 6x + 9 > 0 is x ≠ 3; of ≥ 0 all x; of ≤ 0 only x = 3; of < 0 there is none. For −x² + 2x − 5 < 0 we have D = 4 − 20 < 0 and branches pointing down: the parabola lies below Ox, so the solution is all real numbers.
Class activity

“Sketch contest”: the teacher calls out three coefficients; students draw a sketch in one minute and write the solutions of two inequalities with opposite signs. The owner of the most accurate sketch leads the next round.

Practice
1
Solve x² − 7x + 12 < 0.
2
Solve x² + 4x + 4 ≤ 0.
3
Solve x² − 2x + 5 > 0.
4
Why is −x² + 2x − 3 < 0 true for all x?